I have this stairs' light circuit, but the LED won't turn off. What can I do?
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The capacitor is there to debounce the switch.
When the button is pushed, the negative terminal of the op amp is at 0 volts. The positive terminal is at 2/3 of Vcc. This drives the output of the op amp high (to Vcc).
With the button released the positive terminal stays at 2/3 of Vcc. The negative terminal must rise above that value to drive the output low. Evidently the input impedance of the op amp is less than 2 M ohms so the negative is staying below 2/3 Vcc. (input impedance for an op amp is typically 1 to 2 M ohms). Assuming an input resistance greater than 1 M ohm, decreasing the 1 M ohm resistor to a 500 k ohm resistor would cause the input voltage of the negative input to rise above 2/3 Vcc and the output would shut off. Care must be taken because this will increase the current flow to he input of the op amp which could shorten its life.
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